Union of Sum and Divisors

Let's say there's another solution, $n\neq2017$ , so that $d(n)=2018$ , where $d(n)$ is the sum of the divisors of $n$ . We can immediately observe that $n$ must be composite, as $d(p)=p+1$ for any prime $p$ .

We can also eliminate the possibility that $n$ is a power of a prime. The factors of any power of a prime are just the smaller powers of that prime (including $p^0=1$ ), so $d(p^k) = p^k+p^{k-1}+\cdots +1$ . From this, $d(p^k)-1$ is a multiple of $p$ . But in this case, $d(n)-1=2017$ , which is itself a prime. In other words, $n$ must have more than one prime factor.

A key property of the function $d$ is that it is multiplicative - ie, for any two coprime integers $a$ and $b$ we have $d(ab)=d(a)d(b)$ .

The prime factorisation of $2018$ is $2\cdot1009$ .

Any number with more than one prime factor can be written in at least one way in the form $n=ab$ , with $a$ and $b$ coprime and $b>a>1$ .

Putting all this together, we must have $d(a)=2$ and $d(b)=1009$ . But $d(a)=2$ clearly has no solutions; contradiction. And so there are no such $n$ .

Simple brute force solution using Python 3 .

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for a in range(1, 2018):
    if sum([d for d in range(1, a+1) if a%d==0]) == 2018: print(a)

output : 2017

Since the only output is 2017 the correct answer is $\boxed{No}$

The prime factorization of positive number $N$ can be expressed as $N =\prod_{k=1}^{n} p_k^{e_k}= p_1^{e_1}\cdot p_2^{e_2}\cdots p_{n-1}^{e_{n-1}}\cdot p_n^{e_n}$ where $e_1,e_2,\cdots e_n$ are integer (powers) and here we want to find the sum of all positive divisors such that it is equal to 2018 , ie ; $\prod_{k=1}^{n} \left(\sum_{h=0}^{e_k}p^h_k\right) =\prod_{k=1}^{n}\left(\dfrac{p^{e_k+1}-1}{p_k-1}\right) =1\times 2018=2\times 1009$ Since $2, 1009$ are primes with power 1 each which implies that product cannot proceed more than once or twice ie $n\leq 2$ . Here we can write it as either $\begin{aligned} \left(\dfrac{p^{e_1+1}-1}{p_1-1}\right) & = 2018(*) \\ \text{or} \left(\dfrac{p^{e_1+1}-1}{p_1-1} \right)\left(\dfrac{p^{e_2+1}-1}{p_2-1}\right)&=2\times 1009(**)\end{aligned}$ For the first case $(*)$ we can observe that $\begin{aligned} p^{e_1+1} -2018p & =-2017 \\ p(2018-p^{e_1})& =2017\implies p=2017 \ ,e_1=1 \end{aligned}$ For the second case $(**)$ $\begin{cases} \dfrac{p_1^{e_1+1}-1}{p_1-1}=2\implies p_1(2-p^{e_1})=1\implies \text{no solutions}\\\dfrac{p_2^{e_2+1}-1}{p_2-1}=1009\implies p_2(1009-p_2^{e_2})=1008=2^4\cdot3^2 \cdot 7 \end{cases}$ Solving we obtain $p_2=7$ and $1009-7^{e_2}= 144$ or $7^{e_2}=865\implies e_2 = \dfrac{\log (865)}{7}\not\in\mathbb N$ .

Hence, only the solution that exist , $N=2017$ . Therefore, the answer is No .

For a number n with prime decomposition $n=\prod_k{p_k^{a_k}}$ , the sum of its divisors is $\prod_k{(1+p_k+...+p_k^{a_k})}$ .

Our wanted divisor sum 2018 has prime decomposition 2×1009. Since the factor 2 cannot be expressed as $1+p+...+p^{a}$ , we are looking for single prime $p$ such that $2018=1+p+...+p^a= \frac{p^{a+1}-1}{p-1}$ , and hence $1+2018(p-1) = p^{a+1}$ $\Rightarrow 2018-\frac{2017}{p} = p^{a}$ So we need $p$ to be a prime divisor of $2017$ , which is a prime number itself. Trying $p=2017$ we find $2018=1+p$ , which was given in the problem.We just showed it to be the only solution.